Posted by **Britt** on Monday, November 19, 2012 at 3:33pm.

solve: log8(w-6)=2-log8(w+15)

note: the 8 is to be little at the bottom right corner of log

- Algebra 1 -
**Steve**, Monday, November 19, 2012 at 3:47pm
since 64 = 8^2, then using base-8 logs, we have

log(w-6) = log(64)-log(w+15)

(w-6) = 64/(w+15)

(w-6)(w+15) = 64

w^2 + 9w - 154 = 0

w = 1/2 (-9±√697)

scrap the negative value (why?)

- Algebra 1 -
**Britt**, Monday, November 19, 2012 at 3:49pm
because the answer has got to be a real number which is not a negative number

- Algebra 1 -
**Britt**, Monday, November 19, 2012 at 3:53pm
how do you get 1/2 as a solution?

- Algebra 1 -
**Britt**, Monday, November 19, 2012 at 4:01pm
So technically there would be no solution right?

- Algebra 1 -
**Steve**, Monday, November 19, 2012 at 4:06pm
So, technically, the solution is

x = 1/2 (-9+√697) = 8.70

where does the 1/2 come from? Think back ... back ... back to Algebra I and the quadratic formula.

- Algebra 1 -
**Britt**, Monday, November 19, 2012 at 4:09pm
the quadratic formula is -b+-radical b^2-4ac/a

- Algebra 1 -
**Britt**, Monday, November 19, 2012 at 4:34pm
SO HOW DO YOU GET 1/2 FORM THAT?

- Algebra 1 -
**Reiny**, Monday, November 19, 2012 at 5:26pm
Britt, when Steve and I went to school, the quadratic formula used to be

(-b ± √(b^2 -4ac)/(2a) or

(1/2) (-b ± √(b^2 -4ac)/a , thus the 1/2

I was not aware of any recent changes to the formula.

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